Doubt in Regular Expressions

I started learning Regular Expressions. I am stuck in following.
http://java.sun.com/docs/books/tutorial/extra/regex/literals.html
In the above url, i didnt understand following:
"Note that while the input string is 3 characters long, the start index is 0 and the end index is 3. By convention, ranges are inclusive of the beginning index and exclusive of the end index"
I didnt understand why the end index is 3 instead of 2.

exclusive of the end index means that the end
index is the last index of the match + 1 (one
character after the last index).
Since the last index of the match is 2, then the end
index will be 3.I am finding it hard to understand. Could you please explain in bit more detail. As i am in starting stage, i want to understand it clearly.

Similar Messages

Doubt with Regular Expression (REGEXP_REPLACE)

Hi for All,
how can I get the expected result using regexp_replace?
With Tab As( Select 1 Id, ' .XX ( A => B' TXT From dual Union All
             Select 1 Id, ' , C => 783 );'     TXT From dual )
SELECT REGEXP_REPLACE(Upper(TXT || Lead(TXT)Over(Partition BY id Order BY rownum))
                     ,'(^.*XX *$.+,.+)(\d+)( *$ *;.*$)','\2') AA
FROM tab
AA
3
, C => 783 );
Expected result
AA
783
, C => 783 );Regards,

Hi,
This gets the right results, perhaps for the right reason:
SELECT REGEXP_REPLACE ( Upper(TXT || Lead (TXT) Over (Partition BY id Order BY rownum))
                       , '(^.*XX *$.+,\D+)(\d+)( *$ *;.*$)','\2'
                 )      AS AA
FROM tab
;That is, change the 2nd '.+' to '\D+'.
Consider this regular expression:
SELECT     REGEXP_REPLACE ( 'C => 783'
                 , '(.+)(\d+)'
                 , '\2'
                 )     AS aa
FROM     dual
;Clearly, the 1st argument matches the pattern in the 2nd, but how? All of the interpretations below are equally valid:
\1          \2
C => 78          3
C => 7          83
C =>           783When there is a choice like this, it seems that \1 is made as big as possible.

Doubt in Regular Expressions : java.util.regex

I want to identify words starting with capital letters in a sentence and I want to replace the matched word with "#" added in front of it.... For example, if my input sentence is
"Christopher Williams asked Mike Muller a question"
my output should be,
"#Christopher #Williams asked #Mike #Muller a question"
How do I do that using java.util.regex ?
In perl we can can use *"back references"* in *"replacement string"* . Perl replacement accepts back references whereas java replacement method accepts only strings....
Please help me.....

Your replacement is swallowing the space before the uppercase character, and won't match at the beginning of the line.
Also, it's unnecesarily verbose. String has a replaceAll method (that calls the same methods of Pattern and Matcher under the covers)sentence = s.replaceAll("(^| )([A-Z])", "$1#$2");Disclaimer: I'm no prome, sabre or u/a :-) That can probably be simplified.
db

Doubt with Regular Expression

Hi for All,
how can I get the expected result using regexp_substr or regexp_replace?
with tab as
Select 'TEST( XX_XXXXXX, 12 ) ; AAAA' txt from dual union all
Select 'TEST( AAAAAAAAA , 67 ); 1234' txt from dual union all
Select 'TEST( 92233 ,   47   ); 5234' txt from dual union all
Select 'TEST( AAAAAAAAA , AA ); 897'   txt from dual union all
Select 'TEST( CCCCC 25 );'             txt from dual union all
Select 'TEST CCCCC, 45 );'             txt from dual union all
Select 'TEST( EDCCCC, 45 ;)'           txt from dual union all
Select 'TEST(BBBBBBBBB,12);'           txt from dual
Select regexp_substr(txt, '[^.*,. *][[:digit:]][^\).*$]') from tab
REGEXP_SUBSTR(TXT,'[^.*,.*][[:
12
67
922
897
25
45
45
8 rows selected
Mandatory Criteria
"TEST" And "(" And "," And ")" And ";"
expected result
12
67
47
Null
Null
Null
Null
12Regards,

Something like:
with tab as
Select 'TEST( XX_XXXXXX, 12 ) ; AAAA' txt from dual union all
Select 'TEST( AAAAAAAAA , 67 ); 1234' txt from dual union all
Select 'TEST( 92233 ,   47   ); 5234' txt from dual union all
Select 'TEST( AAAAAAAAA , AA ); 897'   txt from dual union all
Select 'TEST( CCCCC 25 );'             txt from dual union all
Select 'TEST CCCCC, 45 );'             txt from dual union all
Select 'TEST( EDCCCC, 45 ;)'           txt from dual union all
Select 'TEST(BBBBBBBBB,12);'           txt from dual
select txt,
        case
          when regexp_like(txt,'^TEST$.+, *\d+ *$ *;.*$') then regexp_replace(txt,'(^TEST$.+, *)(\d+)( *$ *;.*$)','\2')
        end expected_result
from tab
TXT                          EXPECTED_RESULT
TEST( XX_XXXXXX, 12 ) ; AAAA 12
TEST( AAAAAAAAA , 67 ); 1234 67
TEST( 92233 ,   47   ); 5234 47
TEST( AAAAAAAAA , AA ); 897
TEST( CCCCC 25 );
TEST CCCCC, 45 );
TEST( EDCCCC, 45 ;)
TEST(BBBBBBBBB,12);          12
8 rows selected.
SQL> SY.

Regular expression doubt

hi all
I having input data as SHASHANK:ABA:DEF:GHI:JKL
and output should come as ABA
I tried as below but am getting output as JKL
I need it using regular expressions (am trying to learn how regular expressions can be useful)
please any one help me
SELECT REGEXP_SUBSTR('SHASHANK:ABA:DEF:GHI:JKL' ,'[^:]*$',1) FROM DUALRegards
shashank .k
Edited by: maddy on 10 Jan, 2012 11:28 PM
Edited by: maddy on 10 Jan, 2012 11:28 PM
Edited by: maddy on 10 Jan, 2012 11:28 PM

HI,
Try this..
SELECT REGEXP_SUBSTR ('SHASHANK:ABA:DEF:GHI:JKL', '(ABA)', 1, 1) AS re
FROM DUALThanks,
P Prakash

Regular Expressions and String

How do I return a String array as follow using regular expression.
String[] strArray = {"Now is the time", "you can optionally preview your post","message by using a number of special tokens."}
from this string
<separator>Now is the time</separator><separator>you can optionally preview your post</separator><separator>message by using a number of special tokens.</separator>
Note: The string has the <separator> XML tag

How do I return a String array as follow using regular
expression.
String[] strArray = {"Now is the time", "you can
optionally preview your post","message by using a
number of special tokens."}
from this string
<separator>Now is the time</separator><separator>you
can optionally preview your
post</separator><separator>message by using a number
of special tokens.</separator>
Note: The string has the <separator> XML tag
This cannot be done using simple regular expressions - at least not if your number of <separator>s is random, which is what you seem to imply.
Simple regular expressions are one-off, that means it can have a String array as a result, but only to the amount of brackets in the regex.
a regex like:
<separator>([^<]*)</separator><separator>([^<]*)</separator><separator>([^<]*)</separator>
would return what you want, but I doubt that it is as flexible as you want it to be.

Search with regular expression like this "[^\(]+mytype"

I tried regular expression which find me line which has "mytype" but does not precede with "("
I wrote it like this: [^\(]+mytype
It works perfectly with python reg exp. But it does not work in MSVC 2013 search(ctrl+shift+f). Do you know why? Is it bug?
Is there a way to find variable definition with search function through whole project

Hi Vit,
According to your description, you mean that when you use the
ctrl+shift+f shortcut key to search which line has "mytype" word in the whole project from VS2013 IDE like the following screen shot, but it does not work
in VS2013 IDE, am I right?
If yes, could you please provide me a screen shot about how you search the
mytype?
Generally, when we used the
ctrl+shift+f shortcut key to search, it works fine in VS IDE. Therefore, to check if you install some third-party Add-ins impact your VS IDE, I suggest you can search
the mytype in VS safe mode check if you get same issue.
Reference:
http://msdn.microsoft.com/en-us/library/ms241278.aspx
(1)If you did not get same issue, I doubt that you may be install some Add-ins impact the VS IDE, so I suggest
you can try to disable or un-install them and then check this issue again.
(2)If you still get same issue, I suggest you can try to reset the VS setting file or you can try to repair your
VS check this issue again.
In addition, I suggest you can try to use Ctrl + F shortcut key to search the mytype word.
If I misunderstand your issue, please provide me more detail message about your issue.
Best Regards,
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Problem with my regular expression

Hi
I am trying to find function declarations within source code. I've been trying to use regular expressions, but if anyone else has an easier / better way to do this, as ross perot said, "I'm all ears"
Here's my regular expression printed out:
sb = [.]*void foo[.]*
sb is a stringbuffer variable, so the regex is [.]*void foo[.]*
Can anyone tell me why when I run this pattern on my loaded source file it tells me that the pattern does not match? LOL can you help me fix this?

I doubt that regex is a very good way to find method
declarations, although you can probably make it work
reasonably well for the most common cases.You've got a point. This is just to get the ball rolling. I don't want to have to do full syntax parsing yet.
>
In your particular situation, I'd guess it has to do
with [.]*. I think that inside [], the
dot is a literal dot, not "any character." If you
want "zero or more of anything followed by void" then
it's .*void.
Just tried
sb = .*void foo.*
it still didn't work.
In any case, however, this only covers methods
returning void. For the more general case, I don'tI didn't give you all my code. I actually use a classloader and reflection to determine what the methods are, return values etc. So yes, void in this case is not overly flexible, but I can plug in whatever it is I'm looking for.
think a simple regex will do it. You need a kind of
tool that I can't recall the name of--something to
build and examine an abstract syntax tree perhaps.
Try starting here:
http://www.google.com/search?q=java+abstract+syn
tax+treeI'm not even sure it abstract syntax tree is what you
want, but I think it's at least closer than regex.
There might be something that's a step or two earlier
in the compilation process that will give you what
you want.

Use a regular expression to do this...

I need to process this line : "STRTsomeEND.STRTsome2END.STRTsome3END"
to get an array with this values :
String[] processedLine = {
{"STRTsomeEND."}
{"STRTsome2END."}
{"STRTsome3END."}
I tried to use this regular expression to do the work: (STRT[.*]END.)+ but doesn't work. Also I tried with this expresion STRT[.*]END. But I only got the first ocurrence ("STRTsomeEND") .
Can anybody help me?

zkropotkine wrote:
1) "." is a regex flag.I saw in the table and it said that the "." symbol doesn't need the escape: )
when you ask a question and then someone gives you advice you should
give them the benefit of the doubt (for at least a little while).
did you take a moment to try to understand what i was saying about escaping the "."?
anyway, your regex should be:
(START)(.*?)(END)\ \ .
the "?" after the "\*" means "reluctant". you dont want the ".\*" to grab up everything.
with the reluctant modifier "?" the regex should stop and grab the first available "END".
and finally, if you meant the "." at the end of "STARTsomethingEND." literally then:
you need to escape the "."

Logical AND in Java Regular Expressions

I'm trying to implement logical AND using Java Regular Expressions.
I couldn't figure out how to do it after reading Java docs and textbooks. I can do something like "abc.*def", which means that I'm looking for strings which have "abc", then anything, then "def", but it is not "pure" logical AND - I will not find "def.*abc" this way.
Any ideas, how to do it ?
Baken

First off, looks like you're really talking about an "OR", not an "AND" - you want it to match abc.*def OR def.*abc right? If you tried to match abc.*def AND def.*abc nothing would ever match that, as no string can begin with both "abc" and "def", just like no numeric value can be both 2 and 5.
Anyway, maybe regex isn't the right tool for this job. Can you not simply programmatically match it yourself using String methods? You want it to match if the string "starts with" abc and "ends with" def, or vice-versa. Just write some simple code.

Help in Regular expression

Hello..
I wanted to write a regular expression to match the foll string..
 
 NEW ORLEANS, Louisiana (CNN) 
-- Two years after Hurricane Katrina devastated coastal areas of Louisiana and Mississippi, residents say much of America has forgotten their plight.
 
I tried doing..
Matcher matcher= Pattern.compile(" ([^<^>]+?)", Pattern.CASE_INSENSITIVE).matcher(story);
Its not working...
is there any other soln?

Theres probably a better way to do this but here's a way that works.
import java.util.regex.*;
public class RegexTester{
public static void main(String[] args){
 String text =
 " " +
 " NEW ORLEANS, Louisiana (CNN) " +
 "-- Two years after Hurricane Katrina devastated coastal areas of Louisiana and Mississippi," +
 "residents say much of America has forgotten their plight." +
 " ";
 String regex = ">((?:\\s*[\\S&&[^<>]]+\\s*)*?)<";
 Pattern p = Pattern.compile(regex);
 Matcher m = p.matcher(text);
 while(m.find()){
 System.out.println("Match: '" + m.group(1) + "'");
}

Help in regular expression matching

I have three expressions like
1) [(y2009)(y2011)]
2) [(y2008M5)(y2011M3)] or [(y2009M5)(y2010M12)]
3) [(y2009M1d20)(y2011M12d31)]
i want regular expression pattern for the above three expressions
I am using :
REGEXP_LIKE(timedomainexpression, '???[:digit:]{4}*[:digit:]{1,2}???[:digit:]{4}*[:digit:]{1,2}??', 'i');
but its giving results for all above expressions while i want different expression for each.
i hav used * after [:digit:]{4}, when i am using ? or . then its giving no results. Please help in this situation ASAP.
Thanks

I dont get your question Can you post your desired output? and also give some sample data.
Please consider the following when you post a question.
1. New features keep coming in every oracle version so please provide Your Oracle DB Version to get the best possible answer.
You can use the following query and do a copy past of the output.
select * from v$version 2. This forum has a very good Search Feature. Please use that before posting your question. Because for most of the questions
that are asked the answer is already there.
3. We dont know your DB structure or How your Data is. So you need to let us know. The best way would be to give some sample data like this.
I have the following table called sales
with sales
as
 select 1 sales_id, 1 prod_id, 1001 inv_num, 120 qty from dual
 union all
 select 2 sales_id, 1 prod_id, 1002 inv_num, 25 qty from dual
select *
from sales 4. Rather than telling what you want in words its more easier when you give your expected output.
For example in the above sales table, I want to know the total quantity and number of invoice for each product.
The output should look like this
Prod_id sum_qty count_inv
1 145 2 5. When ever you get an error message post the entire error message. With the Error Number, The message and the Line number.
6. Next thing is a very important thing to remember. Please post only well formatted code. Unformatted code is very hard to read.
Your code format gets lost when you post it in the Oracle Forum. So in order to preserve it you need to
use the {noformat}{noformat} tags.
The usage of the tag is like this.
<place your code here>\
7. If you are posting a *Performance Related Question*. Please read
 {thread:id=501834} and {thread:id=863295}.
 Following those guide will be very helpful.
8. Please keep in mind that this is a public forum. Here No question is URGENT.
 So use of words like *URGENT* or *ASAP* (As Soon As Possible) are considered to be rude.

Regular expression alphabets

Hi
I want to retrieve the data if the data contains a character or a space or '-' thru select query .
Please help me in writing the combination of 3 with regular expression.
Thanks!!

VT wrote:
Hi,
Try this
SELECT *
FROM <TABLE> WHERE REGEXP_LIKE(<COLUMN>, '[a-z -][A-Z -]');cheers
VTThat won't work as it's expecting at least two characters with the first having to be a-z (lower case) or space or "-" followed by A-Z (upper case) or space or "-".
The correct way is either:
[a-zA-Z -]or
[[:alpha:] -]using the alpha set is often preferable as it can work differently with different character sets/languages rather than restricting to just the a-zA-Z ranges.
Generating a reference for your own database characterset/language can be useful...
SQL> select level-1 as asc_code, decode(chr(level-1), regexp_substr(chr(level-1), '[[:print:]]'), CHR(level-1)) as chr,
2 decode(chr(level-1), regexp_substr(chr(level-1), '[[:graph:]]'), 1) is_graph,
3 decode(chr(level-1), regexp_substr(chr(level-1), '[[:blank:]]'), 1) is_blank,
4 decode(chr(level-1), regexp_substr(chr(level-1), '[[:alnum:]]'), 1) is_alnum,
5 decode(chr(level-1), regexp_substr(chr(level-1), '[[:alpha:]]'), 1) is_alpha,
6 decode(chr(level-1), regexp_substr(chr(level-1), '[[:digit:]]'), 1) is_digit,
7 decode(chr(level-1), regexp_substr(chr(level-1), '[[:cntrl:]]'), 1) is_cntrl,
8 decode(chr(level-1), regexp_substr(chr(level-1), '[[:lower:]]'), 1) is_lower,
9 decode(chr(level-1), regexp_substr(chr(level-1), '[[:upper:]]'), 1) is_upper,
10 decode(chr(level-1), regexp_substr(chr(level-1), '[[:print:]]'), 1) is_print,
11 decode(chr(level-1), regexp_substr(chr(level-1), '[[:punct:]]'), 1) is_punct,
12 decode(chr(level-1), regexp_substr(chr(level-1), '[[:space:]]'), 1) is_space,
13 decode(chr(level-1), regexp_substr(chr(level-1), '[[:xdigit:]]'), 1) is_xdigit
14 from dual
15 connect by level <= 256
16 /
ASC_CODE C IS_GRAPH IS_BLANK IS_ALNUM IS_ALPHA IS_DIGIT IS_CNTRL IS_LOWER IS_UPPER IS_PRINT IS_PUNCT IS_SPACE IS_XDIGIT
 0 1
 1 1
 2 1
 3 1
 4 1
 5 1
 6 1
 7 1
 8 1
 9 1 1
 10 1 1
 11 1 1
 12 1 1
 13 1 1
 14 1
 15 1
 16 1
 17 1
 18 1
 19 1
 20 1
 21 1
 22 1
 23 1
 24 1
 25 1
 26 1
 27 1
 28 1
 29 1
 30 1
 31 1
 32 1 1 1
 33 ! 1 1 1
 34 " 1 1 1
 35 # 1 1 1
 36 $ 1 1 1
 37 % 1 1 1
 38 & 1 1 1
 39 ' 1 1 1
 40 ( 1 1 1
 41 ) 1 1 1
 42 * 1 1 1
 43 + 1 1 1
 44 , 1 1 1
 45 - 1 1 1
 46 . 1 1 1
 47 / 1 1 1
 48 0 1 1 1 1 1
 49 1 1 1 1 1 1
 50 2 1 1 1 1 1
 51 3 1 1 1 1 1
 52 4 1 1 1 1 1
 53 5 1 1 1 1 1
 54 6 1 1 1 1 1
 55 7 1 1 1 1 1
 56 8 1 1 1 1 1
 57 9 1 1 1 1 1
 58 : 1 1 1
 59 ; 1 1 1
 60 < 1 1 1
 61 = 1 1 1
 62 > 1 1 1
 63 ? 1 1 1
 64 @ 1 1 1
 65 A 1 1 1 1 1 1
 66 B 1 1 1 1 1 1
 67 C 1 1 1 1 1 1
 68 D 1 1 1 1 1 1
 69 E 1 1 1 1 1 1
 70 F 1 1 1 1 1 1
 71 G 1 1 1 1 1
 72 H 1 1 1 1 1
 73 I 1 1 1 1 1
 74 J 1 1 1 1 1
 75 K 1 1 1 1 1
 76 L 1 1 1 1 1
 77 M 1 1 1 1 1
 78 N 1 1 1 1 1
 79 O 1 1 1 1 1
 80 P 1 1 1 1 1
 81 Q 1 1 1 1 1
 82 R 1 1 1 1 1
 83 S 1 1 1 1 1
 84 T 1 1 1 1 1
 85 U 1 1 1 1 1
 86 V 1 1 1 1 1
 87 W 1 1 1 1 1
 88 X 1 1 1 1 1
 89 Y 1 1 1 1 1
 90 Z 1 1 1 1 1
 91 [ 1 1 1
 92 \ 1 1 1
 93 ] 1 1 1
 94 ^ 1 1 1
 95 _ 1 1 1
 96 ` 1 1 1
 97 a 1 1 1 1 1 1
 98 b 1 1 1 1 1 1
 99 c 1 1 1 1 1 1
 100 d 1 1 1 1 1 1
 101 e 1 1 1 1 1 1
 102 f 1 1 1 1 1 1
 103 g 1 1 1 1 1
 104 h 1 1 1 1 1
 105 i 1 1 1 1 1
 106 j 1 1 1 1 1
 107 k 1 1 1 1 1
 108 l 1 1 1 1 1
 109 m 1 1 1 1 1
 110 n 1 1 1 1 1
 111 o 1 1 1 1 1
 112 p 1 1 1 1 1
 113 q 1 1 1 1 1
 114 r 1 1 1 1 1
 115 s 1 1 1 1 1
 116 t 1 1 1 1 1
 117 u 1 1 1 1 1
 118 v 1 1 1 1 1
 119 w 1 1 1 1 1
 120 x 1 1 1 1 1
 121 y 1 1 1 1 1
 122 z 1 1 1 1 1
 123 { 1 1 1
 124 | 1 1 1
 125 } 1 1 1
 126 ~ 1 1 1
 127 1
 128 Ç 1 1 1
etc.
{code}

Help in query using regular expression

HI,
I need a help to get the below output using regular expression query. Please help me.
SELECT REGEXP_SUBSTR ('PWRPKG(P/W+P/L+CC)', '[^+]+', 1, lvl) val, lvl
FROM DUAL,(SELECT LEVEL lvl FROM DUAL
CONNECT BY LEVEL <=(SELECT MAX ( LENGTH ('PWRPKG(P/W+P/L+CC)') - LENGTH (REPLACE ('PWRPKG(P/W+P/L+CC)','+',NULL))+ 1) FROM DUAL));
I need the output as
correct result:
==============
val lvl
P/W 1
P/L 2
CC 3
But i tried the above it is not coming the above result. Please help me where i did a mistake.
Thanks in advance

Frank gave you a solution in your other thread. You could simplify it if you are on 11g:
SQL> select * from table_x
2 /
TXT
TECHPKG(INTELLI CC+FRT SONAR)
PWRPKG(P/W+P/L+CC)
select txt,
 regexp_substr(
 txt,
 '(.*\()*([^+)]+)',
 1,
 column_value,
 null,
 2
 ) element,
 column_value element_number
from table_x,
 table(
 cast(
 multiset(
 select level
 from dual
 connect by level <= regexp_count(txt,'\+') + 1
 as sys.OdciNumberList
order by rowid,
 column_value
TXT ELEMENT ELEMENT_NUMBER
TECHPKG(INTELLI CC+FRT SONAR) INTELLI CC 1
TECHPKG(INTELLI CC+FRT SONAR) FRT SONAR 2
PWRPKG(P/W+P/L+CC) P/W 1
PWRPKG(P/W+P/L+CC) P/L 2
PWRPKG(P/W+P/L+CC) CC 3
SQL> SY.

Query help in regular expression

Hi all,
SELECT * FROM emp11
WHERE INSTR(ENAME,'A',1,2) >0;
Please let me know the equivalent query using regular expressions.
i have tried this after going through oracle regular expressions documentation.
SELECT * FROM emp11
WHERE regexp_LIKE(ename,'A{2}')
Any help in this regard would be highly appreciated .
Thanks,
P Prakash

please go here
Introduction to regular expressions ...
Thanks,
P Prakash

Doubt in Regular Expressions

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